Integrand size = 25, antiderivative size = 132 \[ \int \frac {x^{11} \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\frac {(c d-b e) x^4}{4 c^2}+\frac {e x^8}{8 c}-\frac {\left (b^2 c d-2 a c^2 d-b^3 e+3 a b c e\right ) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{4 c^3 \sqrt {b^2-4 a c}}-\frac {\left (b c d-b^2 e+a c e\right ) \log \left (a+b x^4+c x^8\right )}{8 c^3} \] Output:
1/4*(-b*e+c*d)*x^4/c^2+1/8*e*x^8/c-1/4*(3*a*b*c*e-2*a*c^2*d-b^3*e+b^2*c*d) *arctanh((2*c*x^4+b)/(-4*a*c+b^2)^(1/2))/c^3/(-4*a*c+b^2)^(1/2)-1/8*(a*c*e -b^2*e+b*c*d)*ln(c*x^8+b*x^4+a)/c^3
Time = 0.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.95 \[ \int \frac {x^{11} \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\frac {2 c (c d-b e) x^4+c^2 e x^8+\frac {2 \left (b^2 c d-2 a c^2 d-b^3 e+3 a b c e\right ) \arctan \left (\frac {b+2 c x^4}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+\left (-b c d+b^2 e-a c e\right ) \log \left (a+b x^4+c x^8\right )}{8 c^3} \] Input:
Integrate[(x^11*(d + e*x^4))/(a + b*x^4 + c*x^8),x]
Output:
(2*c*(c*d - b*e)*x^4 + c^2*e*x^8 + (2*(b^2*c*d - 2*a*c^2*d - b^3*e + 3*a*b *c*e)*ArcTan[(b + 2*c*x^4)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + (-(b* c*d) + b^2*e - a*c*e)*Log[a + b*x^4 + c*x^8])/(8*c^3)
Time = 0.37 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1802, 1200, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^{11} \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx\) |
\(\Big \downarrow \) 1802 |
\(\displaystyle \frac {1}{4} \int \frac {x^8 \left (e x^4+d\right )}{c x^8+b x^4+a}dx^4\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {1}{4} \int \left (\frac {e x^4}{c}+\frac {c d-b e}{c^2}-\frac {\left (-e b^2+c d b+a c e\right ) x^4+a (c d-b e)}{c^2 \left (c x^8+b x^4+a\right )}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-\frac {\text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right ) \left (3 a b c e-2 a c^2 d+b^3 (-e)+b^2 c d\right )}{c^3 \sqrt {b^2-4 a c}}-\frac {\left (a c e+b^2 (-e)+b c d\right ) \log \left (a+b x^4+c x^8\right )}{2 c^3}+\frac {x^4 (c d-b e)}{c^2}+\frac {e x^8}{2 c}\right )\) |
Input:
Int[(x^11*(d + e*x^4))/(a + b*x^4 + c*x^8),x]
Output:
(((c*d - b*e)*x^4)/c^2 + (e*x^8)/(2*c) - ((b^2*c*d - 2*a*c^2*d - b^3*e + 3 *a*b*c*e)*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/(c^3*Sqrt[b^2 - 4*a*c] ) - ((b*c*d - b^2*e + a*c*e)*Log[a + b*x^4 + c*x^8])/(2*c^3))/4
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + ( e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1 )/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.14 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.03
method | result | size |
default | \(-\frac {-\frac {1}{2} c e \,x^{8}+b e \,x^{4}-c d \,x^{4}}{4 c^{2}}+\frac {\frac {\left (-a c e +b^{2} e -c b d \right ) \ln \left (c \,x^{8}+b \,x^{4}+a \right )}{2 c}+\frac {2 \left (a b e -a c d -\frac {\left (-a c e +b^{2} e -c b d \right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{4}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 c^{2}}\) | \(136\) |
risch | \(\text {Expression too large to display}\) | \(2131\) |
Input:
int(x^11*(e*x^4+d)/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
Output:
-1/4/c^2*(-1/2*c*e*x^8+b*e*x^4-c*d*x^4)+1/4/c^2*(1/2*(-a*c*e+b^2*e-b*c*d)/ c*ln(c*x^8+b*x^4+a)+2*(a*b*e-a*c*d-1/2*(-a*c*e+b^2*e-b*c*d)*b/c)/(4*a*c-b^ 2)^(1/2)*arctan((2*c*x^4+b)/(4*a*c-b^2)^(1/2)))
Time = 0.99 (sec) , antiderivative size = 430, normalized size of antiderivative = 3.26 \[ \int \frac {x^{11} \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\left [\frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e x^{8} + 2 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d - {\left (b^{3} c - 4 \, a b c^{2}\right )} e\right )} x^{4} + \sqrt {b^{2} - 4 \, a c} {\left ({\left (b^{2} c - 2 \, a c^{2}\right )} d - {\left (b^{3} - 3 \, a b c\right )} e\right )} \log \left (\frac {2 \, c^{2} x^{8} + 2 \, b c x^{4} + b^{2} - 2 \, a c - {\left (2 \, c x^{4} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{8} + b x^{4} + a}\right ) - {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} d - {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} e\right )} \log \left (c x^{8} + b x^{4} + a\right )}{8 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e x^{8} + 2 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d - {\left (b^{3} c - 4 \, a b c^{2}\right )} e\right )} x^{4} - 2 \, \sqrt {-b^{2} + 4 \, a c} {\left ({\left (b^{2} c - 2 \, a c^{2}\right )} d - {\left (b^{3} - 3 \, a b c\right )} e\right )} \arctan \left (-\frac {{\left (2 \, c x^{4} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} d - {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} e\right )} \log \left (c x^{8} + b x^{4} + a\right )}{8 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}\right ] \] Input:
integrate(x^11*(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="fricas")
Output:
[1/8*((b^2*c^2 - 4*a*c^3)*e*x^8 + 2*((b^2*c^2 - 4*a*c^3)*d - (b^3*c - 4*a* b*c^2)*e)*x^4 + sqrt(b^2 - 4*a*c)*((b^2*c - 2*a*c^2)*d - (b^3 - 3*a*b*c)*e )*log((2*c^2*x^8 + 2*b*c*x^4 + b^2 - 2*a*c - (2*c*x^4 + b)*sqrt(b^2 - 4*a* c))/(c*x^8 + b*x^4 + a)) - ((b^3*c - 4*a*b*c^2)*d - (b^4 - 5*a*b^2*c + 4*a ^2*c^2)*e)*log(c*x^8 + b*x^4 + a))/(b^2*c^3 - 4*a*c^4), 1/8*((b^2*c^2 - 4* a*c^3)*e*x^8 + 2*((b^2*c^2 - 4*a*c^3)*d - (b^3*c - 4*a*b*c^2)*e)*x^4 - 2*s qrt(-b^2 + 4*a*c)*((b^2*c - 2*a*c^2)*d - (b^3 - 3*a*b*c)*e)*arctan(-(2*c*x ^4 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - ((b^3*c - 4*a*b*c^2)*d - (b^4 - 5*a*b^2*c + 4*a^2*c^2)*e)*log(c*x^8 + b*x^4 + a))/(b^2*c^3 - 4*a*c^4)]
Timed out. \[ \int \frac {x^{11} \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\text {Timed out} \] Input:
integrate(x**11*(e*x**4+d)/(c*x**8+b*x**4+a),x)
Output:
Timed out
Exception generated. \[ \int \frac {x^{11} \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^11*(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 1.02 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95 \[ \int \frac {x^{11} \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\frac {c e x^{8} + 2 \, c d x^{4} - 2 \, b e x^{4}}{8 \, c^{2}} - \frac {{\left (b c d - b^{2} e + a c e\right )} \log \left (c x^{8} + b x^{4} + a\right )}{8 \, c^{3}} + \frac {{\left (b^{2} c d - 2 \, a c^{2} d - b^{3} e + 3 \, a b c e\right )} \arctan \left (\frac {2 \, c x^{4} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, \sqrt {-b^{2} + 4 \, a c} c^{3}} \] Input:
integrate(x^11*(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="giac")
Output:
1/8*(c*e*x^8 + 2*c*d*x^4 - 2*b*e*x^4)/c^2 - 1/8*(b*c*d - b^2*e + a*c*e)*lo g(c*x^8 + b*x^4 + a)/c^3 + 1/4*(b^2*c*d - 2*a*c^2*d - b^3*e + 3*a*b*c*e)*a rctan((2*c*x^4 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^3)
Time = 24.71 (sec) , antiderivative size = 8521, normalized size of antiderivative = 64.55 \[ \int \frac {x^{11} \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\text {Too large to display} \] Input:
int((x^11*(d + e*x^4))/(a + b*x^4 + c*x^8),x)
Output:
x^4*(d/(4*c) - (b*e)/(4*c^2)) + (e*x^8)/(8*c) - (log(a + b*x^4 + c*x^8)*(4 *b^4*e + 16*a^2*c^2*e - 4*b^3*c*d + 16*a*b*c^2*d - 20*a*b^2*c*e))/(2*(64*a *c^4 - 16*b^2*c^3)) + (atan((8*c^8*(4*a*c - b^2)^2*(x^4*(((a*c - b^2)*(((( ((((448*b^4*c^10*d - 448*b^5*c^9*e - 384*a*b^2*c^11*d + 832*a*b^3*c^10*e)/ c^8 - (256*b^3*c^4*(4*b^4*e + 16*a^2*c^2*e - 4*b^3*c*d + 16*a*b*c^2*d - 20 *a*b^2*c*e))/(64*a*c^4 - 16*b^2*c^3))*(b^3*e + 2*a*c^2*d - b^2*c*d - 3*a*b *c*e))/(8*c^3*(4*a*c - b^2)^(1/2)) - (32*b^3*c*(b^3*e + 2*a*c^2*d - b^2*c* d - 3*a*b*c*e)*(4*b^4*e + 16*a^2*c^2*e - 4*b^3*c*d + 16*a*b*c^2*d - 20*a*b ^2*c*e))/((4*a*c - b^2)^(1/2)*(64*a*c^4 - 16*b^2*c^3)))*(4*b^4*e + 16*a^2* c^2*e - 4*b^3*c*d + 16*a*b*c^2*d - 20*a*b^2*c*e))/(2*(64*a*c^4 - 16*b^2*c^ 3)) - (((144*b^5*c^8*d^2 + 144*b^7*c^6*e^2 - 240*a*b^3*c^9*d^2 + 96*a^2*b* c^10*d^2 - 528*a*b^5*c^7*e^2 + 480*a^2*b^3*c^8*e^2 - 288*b^6*c^7*d*e + 768 *a*b^4*c^8*d*e - 432*a^2*b^2*c^9*d*e)/c^8 - (((448*b^4*c^10*d - 448*b^5*c^ 9*e - 384*a*b^2*c^11*d + 832*a*b^3*c^10*e)/c^8 - (256*b^3*c^4*(4*b^4*e + 1 6*a^2*c^2*e - 4*b^3*c*d + 16*a*b*c^2*d - 20*a*b^2*c*e))/(64*a*c^4 - 16*b^2 *c^3))*(4*b^4*e + 16*a^2*c^2*e - 4*b^3*c*d + 16*a*b*c^2*d - 20*a*b^2*c*e)) /(2*(64*a*c^4 - 16*b^2*c^3)))*(b^3*e + 2*a*c^2*d - b^2*c*d - 3*a*b*c*e))/( 8*c^3*(4*a*c - b^2)^(1/2)))*(4*b^4*e + 16*a^2*c^2*e - 4*b^3*c*d + 16*a*b*c ^2*d - 20*a*b^2*c*e))/(2*(64*a*c^4 - 16*b^2*c^3)) - (((((((448*b^4*c^10*d - 448*b^5*c^9*e - 384*a*b^2*c^11*d + 832*a*b^3*c^10*e)/c^8 - (256*b^3*c...
\[ \int \frac {x^{11} \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\int \frac {x^{11} \left (e \,x^{4}+d \right )}{c \,x^{8}+b \,x^{4}+a}d x \] Input:
int(x^11*(e*x^4+d)/(c*x^8+b*x^4+a),x)
Output:
int(x^11*(e*x^4+d)/(c*x^8+b*x^4+a),x)